Julie, a third-grade student, believes that the moon disappears on certain days every month. Explain why the following information is or is not scientific evidence which can be used to evaluate Julie’s hypothesis.
A train is travelling east along a straight run of track at 54.0 km/hr. Inside, two siblings 1.90 m apart are playing catch directly across the aisle. The kid wearing a Catherine Wheel T-shirt throws the ball horizontally north. The ball crosses the train and is caught 1.05 s later by her little brother. (Ignore any effects of gravity or friction.)
a) Find the magnitude of the ball's velocity from the little brother's point of view.
b) What is the magnitude of the velocity of the ball as seen by someone standing still outside the train?
c) What is the angle between the direction of the ball and the direction of the train as measured by someone standing still outside the train?
Answer:
A.) 1.81 m/s
B.) 16.07 m/s
C.) 6.88 degree.
Explanation:
Given that Inside, two siblings 1.90 m apart are playing catch directly across the aisle. The kid wearing a Catherine Wheel T-shirt throws the ball horizontally north. The ball crosses the train and is caught 1.05 s later by her little brother.
The velocity of the ball will be
Velocity = displacement/ time
Where displacement = 1.90 m
Time = 1.05 s
Velocity = 1.9 / 1.05
Velocity = 1.8095
Velocity = 1.81 m/s
A.) From the little brother point of view, the magnitude of ball's velocity will be 1.81 m/s
B.) The magnitude of the velocity of the ball as seen by someone standing still outside the train will be achieved by first converting km/h to m/s
(54 × 1000) / 3600 = 15 m/s
Then, use pythagorean theorem to find the velocity
V = sqrt ( 15^2 + 1.81^2)
V = sqrt ( 225 + 3.2761)
V = sqrt ( 258.2761)
V = 16.07 m/s
C.) The angle will be
Tan Ø = 1.81 / 15
Tan Ø = 0.1206666
Ø = tan^-1(0. 1206666)
Ø = 6.88 degree.
What is a population?
Group of answer choices
A. a group of individuals of the same species living in the same area
B. a group of different species living in the same area
C. a group of individuals of the same species living in separated areas a group of different species living in separated areas
Answer:
A
Explanation:
A cement block accidentally falls from rest from the ledge of a 51.1-m-high building. When the block is 13.4 m above the ground, a man, 1.70 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?
Answer:
The man has, at most, 0.4 secs to get out of the way
Explanation:
First, we will determine, the time the block takes to reach height 13.4 m above the ground.
Also, we will determine the time it will take to reach the man, if he does not notice it.
Then, the difference in the two times will be the time the man has to get out of the way.
From the formula
[tex]h =ut - \frac{1}{2}gt^{2}[/tex]
Where [tex]h[/tex] is height
[tex]u[/tex] is the initial velocity
[tex]t[/tex] is time
and [tex]g[/tex] is acceleration due to gravity ( Take [tex]g =[/tex] 9.8 m/s²)
Let the time the block takes to reach height 13.4 m be [tex]t_{1}[/tex]
The cement block falls from a 51.1 m high building, then by the time it reaches a height 13.4 m above the ground, the displacement is
13.4 m - 51.1 m = - 37.7 m
The is the height the block has fallen from by the time it gets to a height 13.4 m above the ground
Then,
[tex]h =ut - \frac{1}{2}gt^{2}[/tex]
[tex]-37.7 = (0)(t_{1}) - \frac{1}{2}(9.8)t_{1}^{2}[/tex]
(NOTE: [tex]u[/tex] = 0 m/s because the cement block falls from rest)
[tex]37.7 = 4.9t_{1}^{2}[/tex]
[tex]t_{1}^{2}= \frac{37.7}{4.9}[/tex]
[tex]t_{1}= \sqrt{7.69} \\[/tex]
[tex]t_{1}= 2.77 secs[/tex]
Hence, It takes the block 2.77 secs to reach height 13.4 m
Also,
Let the time it will take to reach the man be [tex]t_{2}[/tex]
The man is 1.70 m tall, hence, to reach the man, the cement block would have reached a height ( 1.70 m - 51.1 m) = - 49.4 m
Then,
[tex]h =ut - \frac{1}{2}gt^{2}[/tex]
[tex]-49.4 = (0)(t_{2}) - \frac{1}{2}(9.8)t_{2}^{2}[/tex]
[tex]49.4 = 4.9t_{2}^{2}[/tex]
[tex]t_{2}^{2} = \frac{49.4}{4.9}[/tex]
[tex]t_{2}= \sqrt{10.08}[/tex]
[tex]t_{2}= 3.17 secs[/tex]
Hence, it will take 3.17 secs to reach the man
Now, for the time the man has to get out of the way, that is
[tex]t_{2} - t_{1}[/tex] = 3.17 secs - 2.77 secs
= 0.4 secs
Hence, the man has, at most, 0.4 secs to get out of the way
After a group of researchers conducts a new experiment that has never been conducted before, what is the best way for other scientists to make sure that the data is real?
Answer:
I think it would be interviewing the researchers I'm unsure though
Answer:
Probably C) Conducting the experiment again
Explanation:
Option A is wrong because there is a chance that the group of researchers made a mistake or got the wrong data. So when interviewing the researchers, there is a chance of them giving the wrong data, especially since the experiment has never been conducted before.
Option B is wrong because the experiment the researchers where doing was new and has never been conducted before. Therefore there is no older research done on that topic.
Option C is correct because when conducted the experiment again, they can compare their results with the group of researchers. Checking if the data is correct.
Option D is wrong because the group of researchers might have made a mistake when writing down the procedure, leading to false data.
Which principle(s) of the cell
theory shows that all organisms
are somehow similar?
Explanation:
The unified cell theory states that: all living things are composed of one or more cells; the cell is the basic unit of life; and new cells arise from existing cells.
hope it helps
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a golf ball is launched straight up with an initial velocity of 15.68m/s. Calculate the balls displacement and velocity every 0.4 sec into the ball's flight. How long will it take the ball in the air?
Explanation:
Given:
v₀ = 15.68 m/s
a = -9.8 m/s²
t = 0.4n s, n = 0, 1, 2, etc.
Find: Δy and v
To find displacement, use the equation:
Δy = v₀ t + ½ at²
To find velocity, use the equation:
v = at + v₀
[tex]\left\begin{array}{ccc}t&y&v\\0&0&15.68\\0.4&5.49&11.76\\0.8&9.41&7.84\\1.2&11.76&3.92\\1.6&12.54&0\\2.0&11.76&-3.92\\2.4&9.41&-7.84\\2.8&5.49&-11.76\\3.2&0&-15.68\end{array}\right[/tex]
The ball lands after 3.2 seconds.
Answer:
calculated velocities:
(t,v(t))
(0.4,11.756)
(0.8,7.832)
(1.2,3.908)
(1.6,-0.016)
calculated displacements:
[tex](t,\Delta x)[/tex]
(0.4, 5.487)
(0.8,9.4048)
(1.2, 11.7528)
(1.6, 12.5312)
the ball spends 3.2 seconds in the air
Explanation:
Assuming the ball is launched on earth with standard gravity and that the effects of drag is neglected.
assume that 'up' is positive direciton since that is direction of travel
'down' is negative direction
formula for velocity is [tex]v(t)=v_0+at[/tex]
where [tex]v(t)[/tex] is the velocity at time [tex]t[/tex]
[tex]v_0[/tex] is the initial velocity at time [tex]t=0[/tex]
[tex]a[/tex] is the acceleration in the direction of the velocity
velocity is in m/s and accelration is in m/s^2 and t is in seconds
for displacement: [tex]\Delta x=v_0 t+\frac{1}{2}at^2[/tex]
Known stuffs:
in our case, we know we are on earth so, [tex]a=-9.81[/tex]m/s^2. the negative sign indicates that the direction is oposite the velocity since the ball travels up but gravity pulls down
the initial velocity is given as [tex]v_0=15.68[/tex]m/s
therefore, our equation for velocity is
[tex]v(x)=15.68-9.81t[/tex]
now we just need to plot it every 0.4 seconds
this is easy subsituteion. here's the first:
[tex]v(0.4)=15.68-9.81(0.4)=11.756[/tex]m/s
if we keep plotting this, we get the following table
(t,v(t))
(0.4,11.756)
(0.8,7.832)
(1.2,3.908)
(1.6,-0.016) the velocity is negative meaning that the ball is now going 'down' and therefor the ball has just passed the peak of it's upward travel
so since it took about 1.6 seconds to get here, it will spend about twice this time in the air or about 3.2 seconds
to get an exact answer, solve for the time when velocity is 0, ie when the ball has reached its peak and stopped moving
set [tex]v(t)=0[/tex] and solve for t
[tex]0=15.68-9.81t\rightarrow 15.68=9.81t\rightarrow t=1.598s[/tex]
so it takes 1.598 seconds to get to peak, therefore it spends twice this time in the air or 3.197 seconds in the air. it's close enough to 3.2 seconds
the ball spends 3.2 seconds in the air
displacement:
[tex]\Delta x=v_0t+\frac{1}{2}at^2[/tex]
[tex]\Delta x=15.68t+\frac{1}{2}(-9.81)t^2[/tex]
some values are
[tex](t,\Delta x)[/tex]
(0.4, 5.487)
(0.8,9.4048)
(1.2, 11.7528)
(1.6, 12.5312)
An airplane traveled 2,400 meters in 3 seconds. How fast was the plane going?
Answer:
800 miles per hour
Explanation:
_____ electrons are the most important electrons to an atom .
Answer:
stable is the answer
Explanation:
stable electrons
Which refers to the ratio of output work to input work of a machine expressed as a percent?
total work
wattage
efficiency
power
Answer:
Efficiency Or C. on e2020
The ratio of output work to input work of a machine expressed as a percent refers to efficiency of the machine.
What is efficiency of machine?How well a machine converts its input energy into useable output energy, or work, is referred to as its efficiency. It is the portion or percentage of the output divided by the input, and it plays a significant role in a machine's utility.
It denotes the degree of performance that requires the fewest inputs to provide the greatest amount of output in more mathematical or scientific words. It frequently particularly refers to the ability of a particular application of effort to deliver a particular output with the least amount of waste, expense, or superfluous work. Efficiency covers a wide range of inputs and outputs across numerous disciplines and sectors.
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a ball rolling down a hill was displaced 21.9 m while using uniformly accelerating from rest, If the final velocity was 7.14 m/s, what was the rate of acceleration?
Answer:
a = 1.16 m/s²
Explanation:
In order to find the rate of acceleration of the ball, we will use third equation of motion, as follows:
2as = Vf² - Vi²
where,
a = rate of acceleration = ?
s = distance covered by the ball = 21.9 m
Vf = Final Velocity of the ball = 7.14 m/s
Vi = Initial Velocity of the ball = 0 m/s (Since, the ball started from rest)
Therefore,
2(a)(21.9 m) = (7.14 m/s)² - (0 m/s)²
a = (50.97 m²/s²)/(43.8 m)
a = 1.16 m/s²
Physics 1102 Experiment 5 Pre Lab Name_________________________________ Instructor name _________________________ You must show and explain all work in a neat and organized way to receive credit. Please show each step for calculations. YOU MUST TURN IN THIS SHEET. 1. (a) What does electrically neutral mean
Answer:
Explanation:
An atom is said to be electrically neutral, if the said atom contains equal or the same numbers of protons and electrons.
Worthy of note is that most bodies are normally close to being electrically neutral. This is because of the number of electrons on the body being equal to the number of protons on the same body. Charging a body means one would have to transfer electric charge to the body, or from the body. As a result, the number of electrons in the body will no longer be equal to the number of protons on the body.
using keplers 3rd law, determine the distance between the earth and the sun
Answer:
93.615 million or more
Explanation:
The Sun is about 93 million miles from Earth to calculate the distance to a star, astronomers observe it from different places along Earth's orbit around the Sun.
For the system of Problem3.18, determine the amount of energy transfer by heat, in kJ per kg of refrigerant.[Problem3.18A closed, rigid tank contains a two‐phase liquid–vapor mixture of Refrigerant 22 initially at−20°Cwith a quality of 50.36%. Energy transfer by heat into the tank occurs until the refrigerant is at a final pressure of 6 bar. Determine the final temperature, in°C. If the final state is in the superheated vapor region, at what temperature, in°C, does the tank contain only saturated vapor?]
Answer:
35.625° C
Explanation:
See attachment
a) Final temperature is 35.625° C
b) 0°
Suppose you first walk 12.0 m in a direction 200 west of north and then 20.0 m in a direction 40.00 south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position?
Complete Question
The complete question is shown on the first uploaded image
Answer:
the compass direction of the resultant displacement is [tex]\theta =4.7^o [/tex] south of west
Explanation:
Generally using cosine we can obtain the resultant R as follows
[tex]R^2 = A^2 + B^2 -2ABcos(70)[/tex]
=> [tex]R = \sqrt{12^2 + 20^2 - 2(12 ) * (20) cos 70}[/tex]
=> [tex]R = 19.48 \ m[/tex]
We can obtain the direction of the resultant by first using sine rule to obtain angle C as follows
[tex]\frac{A}{sin C} = \frac{R}{sin70 }[/tex]
=> [tex]C= sin ^{-1} [\frac{A * (sin 70)}{R} ][/tex]
=> [tex]C = sin ^{-1} [\frac{20 * (sin 70)}{19.48} ][/tex]
=> [tex]C = 74.7 ^o[/tex]
Then the direction is obtained as
[tex]\theta = C - 70[/tex]
=> [tex]\theta = 74.7 - 70[/tex]
=> [tex]\theta =4.7^o [/tex]
Hence the compass direction of the resultant displacement is [tex]\theta =4.7^o [/tex] south of west
What did Rutherford’s model of the atom include that Thomson’s model did not have?
a nucleus
energy levels
electron clouds
smaller particles
Answer:
Yes the answer is A.) A Nucleus
Explanation:
I took the test
Rutherford’s model of the atom included a nucleus that Thomson’s model did not have, therefore the correct option is A
What are atomic models?There are some models that are used to explain the arrangements of subatomic particles inside the atom based on the atomic theory of atom are known as the atomic models.
There are many types of atomic models proposed in past based on their individual assumptions and the experimentations
There are mainly five most widely accepted atomic models
John Dalton's atomic model which based on the assumption that all the entities of the universe are made up of the fundamental particle that is an atom.
J.J. Thomson's model: Plum Pudding model explains the arrangement of subatomic particles by taking the example of plum pudding in which positive charge is distributed uniformly throughout and the negative charge particles are sprinkled on it in a random fashion.
Rutherford's model explained the existence of a nucleus inside the atom by using the alpha particle bombardment experiment on a thin sheet of gold.
Niels Bohr's model explained the revolving nature of the electrons in the different shells around the nucleus of the atom.
Erwin Schrodinger's model explains the arrangement of the subatomic particles as the electron Cloud Model also known as the Quantum Model.
Rutherford’s model of the atom included a nucleus that Thomson’s model did not have, therefore the correct option is A
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9. Which unit goes with time
kg
K
s
mm
An object is thrown upward at an angle θ above the ground, eventually returning to earth. (a) Is there any place along the trajectory where the velocity and acceleration are perpendicular? If so, where? at the initial position of motion at the highest point in the trajectory at the final position of motion velocity and acceleration are never perpendicular
Answer:
Yes, at the highest point in the trajectory velocity and acceleration are perpendicular.
Explanation:
Let [tex]x-[/tex]axis be the ground and [tex]y-[/tex]axis is the height above the ground. Let the initial velocity of the object is [tex]u[/tex] having angle [tex]\theta[/tex] with respect to ground and the acceleration due to gravity , [tex]g[/tex], is acting in vertically downward direction as shown in figure.
As gravitational force is acting in vertical direction, so, it will not change the horizontal velocity of the object. So, at any instant throughout the projectile motion the [tex]x-[/tex]component of the initial velocity will remailns constant, i.e. [tex]u_x=u \cos \theta[/tex].
While, due gravitational force, [tex]y-[/tex]component of the initial velocity will change. Initially, [tex]u_y=u \sin \theta[/tex] is in vertically upward direction and gravitational force is actiongn vertically downward direction, so, at first [tex]u_y[/tex] will decrease untill it reaches the highest point of the trajectory as shown. At the highest point the vertical component of the velocity [tex]u_y=0[/tex], so, there is only horizontal component of the velocity. i.e [tex]u_x= u \cos\theta[/tex] .
Now, the resultant velocity of the object at the highest point is [tex]u \cos \theta[/tex] which is in the horizontal direction while the acceleration , [tex]g[/tex] , (due to gravily) is actiog in vertically downward direction.
Hence, at the highest point in the trajectory velocity and acceleration are perpendicular.
Somewhere at the greatest point of the trajectory, vector quantities become perpendicular. Considering gravity constantly pushes downward, a projectile's acceleration would always be downward.Furthermore, because the vertical velocity of something like a projectile approaches zero at the height of its trajectory, the first and only portion of the velocity which thus might not be zero seems to be the horizontal factor.
Thus the response above is appropriate.
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When air sinks, it:_________
(a) compresses and warms.
(b) reaches its equilibrium level and then begins to sink.
(c) expands and cools.
(d) forms into clouds.
Answer:
a) compresses and warms.
Explanation:
When air sinks, it falls down and compresses the air close to the surface of the ground, which raises the temperature of the air (since compression results in a rise in temperature). Also, as the air sinks, it prevent the already warm air close to the surface from rising, further increasing its own temperature, warming it up.
Michael threw a football and it traveled for 8 seconds at a velocity of 3 meters per second. How far did the football travel?
A sperm whale can accelerate at about 0.0800 m/s 2 0.0800 m/s 2 when swimming on the surface of the ocean. How far will a sperm whale travel if it starts at a speed of 0.700 m/s 0.700 m/s and accelerates to a speed of 2.18 m/s 2.18 m/s ? Assume the whale travels in a straight line.
Answer:
26.64 m
Explanation:
Given the following :
Acceleration at ocean surface = 0.0800 m/s²
Distance covered if initial speed = 0.700 m/s and accelerates to a speed of 2.18m/s
Using the equation :
v² = u² + 2as
Where ;
v = final velocity ; u = initial velocity ; a = acceleration ; s = distance covered
Therefore,
v² = u² + 2as
2.18² = 0.7² + (2 × 0.08 × s)
4.7524 = 0.49 + 0.16s
4.7524 - 0.49 = 0.16s
4.2624 = 0.16s
s = 4.2624 / 0.16
s = 26.64 m
A box of mass 20kg is pulled up an inclined plane by a force of 285N. Given that the value of the incline angle is 30 degrees and the coefficient of dynamic friction is 0.72, what is the speed with which the box is moving with, assuming it takes 4seconds to reach the top of the incline?
Given :
Mass of box , m = 250 kg.
Force applied , F = 285 N.
The value of the incline angle is 30°.
the coefficient of dynamic friction is [tex]\mu=0.72[/tex] .
To Find :
The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.
Solution :
Net force applied in box is :
[tex]F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N[/tex]
Acceleration , [tex]a=\dfrac{F}{m}=\dfrac{60.29}{20}=3.01\ m/s^2[/tex].
By equation of motion :
[tex]v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s[/tex]
Therefore, the speed of box is 12.04 m/s.
Hence, this is the required solution.
